TITLE: Element of a group G raise to the index of a subgroup H is in H
QUESTION [11 upvotes]: Let $H$ be normal in $G$ and $[G:H]=m$. Prove that $x^m$ is in $H$ for
  all $x$ in $G$.

Since $H$ is normal its left cosets form a group. Also since $m$ is the order of that group $(xH)^m=x^mH=H$, which implies that $x^m \in H$.
I am actually not sure what my question is, it just hard to believe that if I pick any element of a group $G$ and I raise it to $m$ the resulting element will be in $H$. Anyways, is there another way one prove this? Thanks.

REPLY [13 votes]: It's only a superficial change, but since you seem to be having issues accepting your (correct) argument, maybe you will like this better:
Instead of talking (explicitly) about cosets, let's talk (equivalently, of course!) about the quotient group and the quotient map $q: G \rightarrow G/H$.  For $x \in G$, since $\# (G/H) = m$, by Lagrange's Theorem, 
$q(x)^m = e$ (the identity) in $G/H$.  But since $q$ is a homomorphism, $e = q(x)^m = q(x^m)$, so $x^m \in H$.
In any case: believe it!  It's an extremely useful result: I have used it many times in my work in algebra and number theory and have several times noted that students don't seem to be as aware of it as they should be.  (In fact, I like to think of this in terms of the period-index problem in Galois cohomology, but that's really beyond the scope of this answer.)